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July 31st, 2015, 01:17 AM   #1
Joined: Jun 2015
From: South Africa

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Symmetry properties of graphs

In each of the following cases find the least positive value of α for which:
a) cos(α-θ)˚=sinθ˚
b) sin(α-θ)˚=cos(α+θ)˚
There are a whole bunch more...
I did a) easily because obviously cos(90-θ)˚=sinθ˚, so α = 90.

I'm struggling to work out b) though...
Thanks for any help.
Jmun is offline  
July 31st, 2015, 04:50 PM   #2
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This is how I would do it.
Simplifying $\sin(\alpha - \theta) = \cos(\alpha + \theta)$, we get
\sin(\alpha - \theta) = \cos(\alpha + \theta)\\
\sin\alpha\cos\theta - \sin\theta\cos\alpha = \cos\alpha\cos\theta - \sin\alpha\sin\theta\\
\sin\alpha(\cos\theta + \sin\theta) = \cos\alpha(\cos\theta + \sin\theta)\qquad\qquad\qquad(1)
In cases where $\cos\theta + \sin\theta = 0$, equality holds trivially.
Otherwise, $(1)$ can be simplified to
$$\sin\alpha = \cos\alpha,$$
$$\tan\alpha = 1.$$
We now see that the answer is $\alpha = 45^\circ$
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